Questions (re-rcad introductory remarks): pKa_2 + pKb_2 should add up to 14. Why

Question

Questions (re-rcad introductory remarks): pKa_2 + pKb_2 should add up to 14. Why? (The answer is NOT because 7 + 7 = 14!) pKa_1 + pKb_1 should NOT add up to 14. Why? Provide an explanation for each of the scenarios that follow. How will the equivalence point volumes and Ka values for the acid be affected if the following were to occur? You must provide an explanation for how you arrived at your conclusion. The NaOH is actually 0.150 M, not 0.100 M as labeled. Your buret has some water in it before adding NaOH solution. Your pipet has some water in it before you pipct the solution of the acid.

Explanation / Answer

Q.6) Pka + pKb = 14, Why or Why not ?

Answer-6)

Answer for this question is related to the Ionic product of water at various temperature.

Explanation,

I) Water is an ampholyte i.e. behave as an acid (by donating H+) and behave as base (by accepting H+). This behavior of water is due to its autoionization as,

H2O <-----------> H+ + OH–

Dissociation constant (K) for water autoionization is,

K = [H+][OH–]/[H2O]

As H2O ionizes only slightly therefor [H2O] = constant

Hence,

K = [H+][OH–]/[H2O] /constant

K x constant = [H+][OH–]

So,

Kw = [H+][OH–] ………………… (where Kw = K x constant) ……………(1)

This is the ionic product of water and at constant temperature 25 oC its value is Kw = 1 x 10-14.

But at temperature other than 25oC, Kw 1 x 10-14.

Now consider an ionization of weak acid HA with dissociation constant Ka,

HA + H2O <------------> H3O+ + A–

Equilibrium constant Ka can be given as

Ka = [A–][H3O+]/[HA]   ………………. ([H2O] is almost constant)   ……………….(2)

And for ionization of Base A– with dissociation constant Kb,

A– + H2O <-----------> AH + OH–

Equilibrium constant Kb can be given as,

Kb = [AH][OH–] / [A–]      ……………….. ([H2O] is almost constant ………………(3)

Then,

Ka x Kb = ([A–][H3O+]/[HA] ) x ([AH][OH–] / [A–])

Ka x Kb = [H3O+] [OH–] .................... (like terms canceelled out from Nm and Dm)

Ka x Kb = Kw ………………………… (At 25 oC)

Ka x Kb = 1x 10-14

Taking –log of both sides,

–log (Ka x Kb) = –log (1x 10-14)

(–log Ka) + (–log Kb) = 14

pKa + pKb = 14 ………………(at 25 oC)

But as at temperature other than 25 oC Kw 1 x 10-14

pKa + pKb 14 ………………. (at temperature other than 25 oC)

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